If $x, y$ are two positive integers such that $x+y=20$ and the maximum value of $x^3 y$ is $k$ at $x=\alpha, y=\beta$ then $\frac{k}{\alpha^2 \beta^2}=$

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Aicharya · Accepted Answer

$x+y=20$ $\frac{\frac{x}{3}+\frac{x}{3}+\frac{x}{3}+y}{4} \geq\left[\left(\frac{x}{3}ight)^3 yight]^{1 / 4} \Rightarrow 5 \geq\left(\frac{x^3 y}{3^3}ight)^{\frac{1}{4}}$ $\Rightarrow \frac{x^5 y}{27} \leq 625 \Rightarrow x^3 y \leq 27 	imes 625 \Rightarrow k=27 	imes 625$ at $\frac{x}{3}=y \Rightarrow 4 y=20 \Rightarrow y=5=\beta, x=15=\alpha$ $\frac{k}{\alpha^2 \beta^2}=\frac{27 	imes 625}{15^2 	imes 5^2}=3=\frac{\alpha}{\beta}$

If $x, y$ are two positive integers such that $x+y=20$ and the maximum value of $x^3 y$ is $k$ at $x=\alpha,…

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